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LMT
2023 LMT Fall
4A
4A
Part of
2023 LMT Fall
Problems
(1)
2023 Fall Theme p4A
Source:
12/23/2023
Let Revolution
(
x
)
=
x
3
+
U
x
2
+
S
x
+
A
(x) = x^3 +Ux^2 +Sx + A
(
x
)
=
x
3
+
U
x
2
+
S
x
+
A
, where
U
U
U
,
S
S
S
, and
A
A
A
are all integers and
U
+
S
+
A
+
1
=
1773
U +S + A +1 = 1773
U
+
S
+
A
+
1
=
1773
. Given that Revolution has exactly two distinct nonzero integer roots
G
G
G
and
B
B
B
, find the minimum value of
∣
G
B
∣
|GB|
∣
GB
∣
.Proposed by Jacob Xu Solution.
392
\boxed{392}
392
Notice that
U
+
S
+
A
+
1
U + S + A + 1
U
+
S
+
A
+
1
is just Revolution
(
1
)
(1)
(
1
)
so Revolution
(
1
)
=
1773
(1) = 1773
(
1
)
=
1773
. Since
G
G
G
and
B
B
B
are integer roots we write Revolution
(
X
)
=
(
X
−
G
)
2
(
X
−
B
)
(X) = (X-G)^2(X-B)
(
X
)
=
(
X
−
G
)
2
(
X
−
B
)
without loss of generality. So Revolution
(
1
)
=
(
1
−
G
)
2
(
1
−
B
)
=
1773
(1) = (1-G)^2(1-B) = 1773
(
1
)
=
(
1
−
G
)
2
(
1
−
B
)
=
1773
.
1773
1773
1773
can be factored as
32
⋅
197
32 \cdot 197
32
⋅
197
, so to minimize
∣
G
B
∣
|GB|
∣
GB
∣
we set
1
−
G
=
3
1-G = 3
1
−
G
=
3
and
1
−
B
=
197
1-B = 197
1
−
B
=
197
. We get that
G
=
−
2
G = -2
G
=
−
2
and
B
=
−
196
B = -196
B
=
−
196
so
∣
G
B
∣
=
392
|GB| = \boxed{392}
∣
GB
∣
=
392
.
2023
FAlL
theme
alg